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kotegsom [21]
3 years ago
14

Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, th

ey push off against one another. Adolf has a mass of 100 kg, and Ed has a mass of 81 kg. Following the push, Adolf swings upward to a height of 0.51 m above his starting point. To what height above his own starting point does Ed rise?
Physics
1 answer:
exis [7]3 years ago
5 0

Answer:

0.78 m

Explanation:

By the conservation of energy, the energy that they gain from potential energy, must be equal to the kinetic energy. So, for Adolf:

Ep = Ek

ma*g*ha = ma*va²/2

Where ma is the mass of Adolf, g is the gravity acceleration (10 m/s²), ha is the height that he reached, and va is the velocity. So:

100*10*0.51 = 100*va²/2

50va² = 510

va² = 10.2

va = √10.2

va = 3.20 m/s

Before the push, both of them are in rest, so the momentum must be 0. The system is conservative, so the momentum after the push must be equal to the momentum before the push:

ma*va + me*ve = 0, where me and ve are the mass and velocity of Ed. So:

100*3.20 + 81ve = 0

81ve = 320

ve = 3.95 m/s

By the conservation of energy for Ed:

me*g*he = me*ve²/2

81*10*he = 81*(3.95)²/2

810he = 631.90

he = 0.78 m

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steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

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